2020-05-16, 16:38 | #1 |
"Rashid Naimi"
Oct 2015
Out of my Body
2^{3}·3^{2}·5^{2} Posts |
Unity Working for any Factorial
With no fractions along the way. Every intermediary value after division is coprime to the factorial, regardless of its size.
Please see the image. Last fiddled with by a1call on 2020-05-16 at 16:48 |
2020-05-16, 17:15 | #2 |
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
9021_{10} Posts |
...with no factorials along the way. either.
Just put any number, integer, real, imaginary, or "x"! Truly outstanding! |
2020-05-16, 17:33 | #3 |
"Rashid Naimi"
Oct 2015
Out of my Body
3410_{8} Posts |
Well if it's not a factorial the no fraction part won't hold (or the coprime part). If you don't mind that you can replace the 2's and the 3's with any value.
Only Factorials can ensure both conditions. No other established notation would do the same. Not even primorials. ETA With the notable exception for 0 which is not a factorial for any number. ETA II Then again, I am not sure if any number can be considered coprime to 0. ETA III Hence, the factorial will have to be greater than 3!. ETA IV Make that greater than 4! Last fiddled with by a1call on 2020-05-16 at 18:27 |
2020-05-16, 18:50 | #4 |
"Rashid Naimi"
Oct 2015
Out of my Body
2^{3}·3^{2}·5^{2} Posts |
I think we better change it to => 9!, Since the valuation of 2 and 3 would have to be greater than 1 in the factorial.
This also means that any n= 2^a* 3^b*m For integers a, b and m where they are all greater than 1 would also work instead of the factorial. But that's not as interesting a notation as a factorial. Last fiddled with by a1call on 2020-05-16 at 18:51 |
2020-05-16, 19:26 | #5 |
"Rashid Naimi"
Oct 2015
Out of my Body
2^{3}·3^{2}·5^{2} Posts |
Here is a nice little spin-off:
Take any factorial n! Find a PRP in the format: p = (n! +/- k)/k Where for largest prime factor q of k you have q <= n And you have found a provable prime p since all the prime factors of p -/+ 1 will be less than or equal to n. Here is a 10312 dd random example: (3338!-230)/230 ETA Would be very much interested to know if there are any Top-5k primes of this format or what is their categorization. Thanks in advance to axn. Last fiddled with by a1call on 2020-05-16 at 19:36 |
2020-05-17, 13:38 | #6 |
"Rashid Naimi"
Oct 2015
Out of my Body
2^{3}×3^{2}×5^{2} Posts |
A search for "!" (Without the quotes) in the Top-5k database only returns factorial-primes with k = 1. So I assume we can conclude that there are no primes of the format with k>1 in the Top-5k.
Furthermore a Google search for "extended factorial primes" (with the quotes) returns 0 results. So I think it is safe to assume that these primes have not been extensively researched. Some more "outstanding" (with the quotes) facts about the format is that primes of the form (n!-k)/k and (n!+k)/k with equal n's & k's are 2 apart and are thus twin primes. Now while "factorial primes" (with the quotes) can never be twin-primes (with the notable exception for 3! & 2!), there are no such restrictions on the extended-factorial-primes. I think this is quite an important concept. Any thoughts on why? Last fiddled with by a1call on 2020-05-17 at 13:53 |
2020-05-17, 15:24 | #7 |
"Rashid Naimi"
Oct 2015
Out of my Body
11100001000_{2} Posts |
Here is a relevant discussion for reference:
https://math.stackexchange.com/quest...al-twin-primes FTR I don't think that perhaps it should not be an "open question" (with quotes). Last fiddled with by a1call on 2020-05-17 at 15:46 |
2020-05-17, 15:58 | #8 |
"Rashid Naimi"
Oct 2015
Out of my Body
2^{3}·3^{2}·5^{2} Posts |
On yet another spin-off proving that there exists a distinct twin prime of the form (n!+/-k)/k for an integer n greater than any given integer will prove the infinitude of twin primes.
ETA DHS-100-A For any given factorial n! greater than 3!, We know that there exists at least one integer k such that (n! +/- k)/k are both coprime to n!. This is because any k whose largest prime factor is equal to or less than n and whose valuations of all its prime factors are less than the valuations of the same prime factors in the n! will satisfy this condition. To-be-continued. Last fiddled with by a1call on 2020-05-17 at 16:50 |
2020-05-17, 19:02 | #9 |
"Rashid Naimi"
Oct 2015
Out of my Body
3410_{8} Posts |
To clarify the
n! = 2^8 * 3^4 * 5^2 * 7^1 * 11^1 Now for any k = 2^a * 3^b * 5^c where 0 <= a < 8, 0 <= b < 4, 0 <= c < 2, for example for k = 2^3 *5^1 = 40, Both (11! +/- 40)/40 will definitely be coprime to 11! and will be 2 apart. So we could say they are twin coprimes to 11!, To be continued... Last fiddled with by a1call on 2020-05-17 at 19:06 |
2020-05-18, 14:53 | #10 |
"Rashid Naimi"
Oct 2015
Out of my Body
3410_{8} Posts |
FTR As of the time of this posting a Google-Search for "Extended Factorial Primes" (with quotes) returns no results. Not even this thread which is unusual.
As of the time of this posting a Google-Search for "Generalized Factorial Primes" (with quotes) returns no results. DHS-150-A For any given factorial n! greater than 2!, let v be the valuation of 2 in n!. Then we know that (n! +/- -2^v)/(2^v) will both have valuations of 2 that are greater than 0. In fact one will definitely have a valuation equal to 1 while the other will definitely have a valuation greater than 1. Other than for 2 the expressions will have no other common prime factors with n!. For example: valuation(19!,2) = 16 valuation((19!+2^16)/(2^16),2) = 1 valuation((19!-2^16)/(2^16),2) = 3 gcd((19!-2^16)/(2^16),19!) = 2^3 gcd((19!+2^16)/(2^16),19!) = 2^1 To-be-continued ... Last fiddled with by a1call on 2020-05-18 at 14:56 |
2020-05-19, 03:12 | #11 |
"Rashid Naimi"
Oct 2015
Out of my Body
3410_{8} Posts |
DHS-160-A For any given factorial n! greater than 2!, let v be the valuation of prime factor 3 of n!.
Then we know that one of the (n! +/- 3^v)/(3^v) will have a valuations of 3 that is 0 and the other will have a valuations of 3 that is greater than 0. Other than for 3 (for exactly one of the pair) the expressions will have no other common prime factors with n!. 2 and/or 3 are the only prime factors that if present (individually or in combination) in k while having a valuation equal to that of their respective valuation in n! (individually or in combination), will never yield both (n! +/- k)/k coprime to n!. To-be-continued ... Last fiddled with by a1call on 2020-05-19 at 03:19 |
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